Problem: Suppose we wanted to evaluate the double integral $S = \iint_D y \, dx \, dy$ by first applying a change of variables from $D$ to $R$ : $\begin{aligned} x &= X_1(u, v) = u - v \\ \\ y &= X_2(u, v) = v^2 + u^3 \end{aligned}$ What is $S$ under the change of variables? Assume $v > 0$. If you know an expression within absolute value is non-negative, do not use absolute value at all. $S = \iint_R $ $du \, dv$
Answer: If we have a transformation $\bold{X} : R \to D$, then we can rewrite an integral under the change of variables: $ \iint_D f(x, y) \, dA = \iint_R f(\bold{X}(u, v)) | J(\bold{X}) | \, du \, dv$ First we need to find the absolute value of the Jacobian, $|J(\bold{X})|$. $\begin{aligned} |J(\bold{X})| &= \left| \det \begin{pmatrix} \dfrac{\partial X_1}{\partial u} & \dfrac{\partial X_1}{\partial v} \\ \\ \dfrac{\partial X_2}{\partial u} & \dfrac{\partial X_2}{\partial v} \end{pmatrix} \right| \\ \\ &= \left| \det \begin{pmatrix} 1 & -1 \\ \\ 3u^2 & 2v \end{pmatrix} \right| \\ \\ &= \left| 2v + 3u^2 \right| \end{aligned}$ We are given that $v > 0$, so $\left| 2v + 3u^2 \right|$ simplifies to $2v + 3u^2$ because $3u^2$ is always positive. Now we substitute $u$ and $v$ in $f(x, y)$. $\begin{aligned} f(x, y) &= f(\bold{X}(u, v)) \\ \\ &= X_2(u, v) \\ \\ &= v^2 + u^3 \end{aligned}$ Putting everything together, we get the integral under the change of variables: $ \iint_R (2v + 3u^2) (v^2 + u^3) \, du \, dv$